Math Problem Statement

Solve the following logarithmic equation.

log Subscript 3 Baseline left parenthesis x plus 11 right parenthesis plus log Subscript 3 Baseline left parenthesis x plus 3 right parenthesis equals 2log3(x+11)+log3(x+3)=2

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Part 1

Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

The solution set is

StartSet nothing EndSetenter your response here.

​(Simplify your answer. Type an exact answer. Use a comma to separate answers as needed.

B.

There is no solution.

Solution

We are given the logarithmic equation:

log3(x+11)+log3(x+3)=2\log_3 (x + 11) + \log_3 (x + 3) = 2

Step 1: Apply the logarithmic property

We can use the property of logarithms that states:

logb(a)+logb(c)=logb(ac)\log_b (a) + \log_b (c) = \log_b (a \cdot c)

Using this property:

log3((x+11)(x+3))=2\log_3 ((x + 11)(x + 3)) = 2

Step 2: Convert the logarithmic equation to an exponential equation

The equation now becomes:

log3((x+11)(x+3))=2\log_3 ((x + 11)(x + 3)) = 2

We can rewrite this in exponential form. Since logb(a)=c\log_b (a) = c implies bc=ab^c = a, we get:

(x+11)(x+3)=32(x + 11)(x + 3) = 3^2

(x+11)(x+3)=9(x + 11)(x + 3) = 9

Step 3: Expand the left-hand side

Expand the left-hand side of the equation:

(x+11)(x+3)=x2+3x+11x+33=x2+14x+33(x + 11)(x + 3) = x^2 + 3x + 11x + 33 = x^2 + 14x + 33

So the equation becomes:

x2+14x+33=9x^2 + 14x + 33 = 9

Step 4: Simplify the equation

Subtract 9 from both sides:

x2+14x+339=0x^2 + 14x + 33 - 9 = 0

x2+14x+24=0x^2 + 14x + 24 = 0

Step 5: Solve the quadratic equation

We solve the quadratic equation:

x2+14x+24=0x^2 + 14x + 24 = 0

We can use the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For the equation x2+14x+24=0x^2 + 14x + 24 = 0, the coefficients are a=1a = 1, b=14b = 14, and c=24c = 24. Plugging these values into the quadratic formula:

x=14±1424(1)(24)2(1)x = \frac{-14 \pm \sqrt{14^2 - 4(1)(24)}}{2(1)}

x=14±196962x = \frac{-14 \pm \sqrt{196 - 96}}{2}

x=14±1002x = \frac{-14 \pm \sqrt{100}}{2}

x=14±102x = \frac{-14 \pm 10}{2}

Now, calculate the two possible values for xx:

x=14+102=42=2x = \frac{-14 + 10}{2} = \frac{-4}{2} = -2 x=14102=242=12x = \frac{-14 - 10}{2} = \frac{-24}{2} = -12

Step 6: Check for extraneous solutions

Since the logarithm is undefined for negative arguments, we must check whether these values of xx make the expressions inside the logarithms positive.

  • For x=2x = -2, x+11=9x + 11 = 9 and x+3=1x + 3 = 1, which are both positive, so this solution is valid.
  • For x=12x = -12, x+11=1x + 11 = -1, which is negative, so this solution is not valid.

Final Answer

The only valid solution is x=2x = -2.

The solution set is:

{2}\{-2\}

The correct choice is A.


Would you like more details on any part of the process?

Here are 5 questions related to this topic:

  1. How do you solve logarithmic equations that involve products?
  2. Can all logarithmic equations be solved by converting to exponential form?
  3. How do you determine whether a logarithmic solution is extraneous?
  4. What happens if the argument of a logarithmic function is negative?
  5. How does the quadratic formula help in solving logarithmic equations?

Tip: Always check for extraneous solutions when solving logarithmic equations, as the logarithm of a negative number is undefined.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Algebra
Quadratic Equations
Exponential Equations

Formulas

log_b(a) + log_b(c) = log_b(a * c)
log_b(a) = c implies b^c = a
Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a

Theorems

Logarithmic properties
Quadratic formula

Suitable Grade Level

Grades 9-12